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Analytical Geometry
1. A variable straight line, drawn through the point of intersection of the straight lines x/a + y/b = 1 and
x/b + y/a = 1 meets the coordinate axes in A and B.
Show that the locus of mid point of AB is the curve
2(a + b)xy = ab(x + y).
2. If sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is
a) square b)circle
c)straight line d)two intersecting lines
3. The line ( p + 2q)x + (p-3q)y = p-q for different values of p and q passes through rhe point
a) (3/2 , 5/2) b) (2/3 , 2/5)
c) (3/5 , 3/5) d) (2/5 , 3/5 )
4. If x/c + x/d = 1 be any line through the intersection of lines x/a + x/b = 1 and x/b + x/a = 1, then
a) 1/a + 1/d = 1/b + 1/c b) 1/b + 1/d = 1/c + 1/a
c) 1/c + 1/d = 1/a + 1/b d)n.o.t
5. The eqn. to the pair of straight lines through the origin and perpendicular
to 2
- 4xy - 7
= 0 is ...................................?
6. If a
+ 2hxy + b
+ 2gx + 2fy + c = 0 represents a pair of lines, equidistant from the origin, then
a) 
= c( bf - ag ) b) 
- 
= c( b
- a
)
c) 
+ 
= c( b
+ a
) d) n.o.t
7. The eqn. of bisector of that angle b/w the lines x + y + 1 = 0 and
2x - 3y - 5 =0 which contains the point ( 10 , -20 ) is ..............................?
8. A point moves so that sum of squares of its distances from the four sides of a square is constant. This point lies on a
a) straight line b) circle
c) ellipse d) parabola
9. Given the system of straight lines a( 2x + y - 3 ) + b( 3x + 2y - 5 ) = 0, the
line of the system situated farthest from the point ( 4, -3 ) has the eqn. ................................?
10. If point (4 , 5 ) is rotated by 
anticlockwise though the origin the new point is ..............................?
11. All points lying inside the triangle formed by the points ( 1,3 ) , ( 5 , 0 ) and ( -1 , 2 ) satisfy
a) 3x + 2y 
0 b) 2x + y - 13 
0
c) 2x - 3y - 12 
0 d) -2x + y 
0
12. The line L has intercepts a and b on the co-ordinate axes. When keeping the origin fixed, the co-ordinate axes are rotated through a fixed angle, then the same line has intercepts p and q on the rotated axes, then
a) 
+ 
= 
+ 
b) 
+ 
= 
+ 
c) 
+ 
= 
+ 
d) 
+ 
= 
+ 
13. The area of triangle is 5sq. units and two of its vertices are(2 ,1 ) and
( 3 , -2 ). The third vertex which lies on yhe line y = x + 3 is
a) (-3/2,3/2) correct b) (5/2,11/2)
c) (7/2.13/2) correct d) (0,0)
14. If one of the diag0nals of a square is along the line x = 2y and one of its vertices is (3,0), then the eqns. of the sides through this vertex are given
by ...........................?
15. The eqn. h
+ ( 1- 
)xy - h
= 0 represents a pair of straight lines, then
a) h =2 necessarily b) h = -2 necessarily
c) h may be any real number d) n.o.t
Comments (11)
for the 15th question..............................
if they should reperesnt a pair of lines then they shud satisfy....
h square greater than ab......caluculate tat and u find is satisfies for all real values of h
for 13th the answer is c draw the traingle and find the length of its base....then height to caluculate the area will be perpendicular distance from the third vertex to the line u have got.....so multiply 1/2 base*height and u will get it
![1) \mbox{Point of intersection = } [\frac{ab}{a+b},\frac{ab}{a+b}] \\ \\<br/>\mbox{Let the equation of line be }->y=mx + c \\ \\<br/>\mbox{Now since it passes through} \frac{ab}{a+b},\frac{ab}{a+b} ,\mbox{ we get relation as }-> \\ \\<br/>c = \frac{[1-m]ab}{a+b} \\ \\<br/>Also , A -> \frac{-c}{m}, 0 \\ \\<br/>B -> 0, c \\ \\<br/>Mid-Point -> x=\frac{-c}{2m} , y=\frac{c}{2} \\ \\<br/>\mbox{which implies }m = \frac{-y}{x} \\ \\<br/>Hence -> \\ \\<br/>y=\frac{c}{2}=\frac{[1-m]ab}{2(a+b)}=\frac{[x+y]ab}{2x(a+b)}\\ \\<br/>or ->\\ \\<br/>\mbox{2xy(a+b)=ab(x+y)}](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/f/6/0f6b73553a26b6e4929849b7459f89d46998f4be.gif)
For 13th question,take the third vertex as (x , x+3) and substitute in the area of triangle formula
[y = x+3 is given]..on solving |4x-4| = 10 will be obtained
4x-4 = 10 4x-4 = -10
x = 7/2 x = -3/2
substitute x=7/2 in y=x+3 substitute x= -3/2 in y= x+3
y = 13/2 y = 3/2
3rd vertex is ( 7/2,13/2) 3rd vertex is (-3/2,3/2)
The questions 1,3,4,5,13,15 are already answered.
Let me answer some of the remaining questions.
2)Let us assume the lines to be X,Y axes for convenience.Then,P(x0,y0) belongs to the locus iff Ix0I+Iy0I=1.Thus the locus is a square formed by the lines IxI+IyI=1 with centre as origin.
2x2+2y2-2ax-2ay+2a2-k=0.
a2.
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for d 5th q
d ansewr is 7x^2+4xy-2y^2=0
c if ther's ne combined equation of straigth lines ax^2-2hxy+by^2=0.....which passes thru d origin then d equation of line perpendicular 2 it is given by bx^2+2hxy+ay^2=0
yeh raata mar lo..
.....wat u hav 2 do is just change d sign of d middle term and change d coefficients of d other two...
note .this is only and only if they pass thru origin